How to fix "IndexError: tuple index out of range" in python?

How to fix "IndexError: tuple index out of range" in python?

我正在使用 sklearn 模块来寻找最合适的模型和模型参数。但是,我在下方遇到意外的索引错误:

> IndexError                                Traceback (most recent call
> last) <ipython-input-38-ea3f99e30226> in <module>
>      22             s = mean_squared_error(y[ts], best_m.predict(X[ts]))
>      23             cv[i].append(s)
> ---> 24     print(np.mean(cv, 1))
> IndexError: tuple index out of range

我想做的是找到最合适的回归量及其参数,但我遇到了上述错误。我调查了 SO 并尝试了 this solution 但仍然出现了同样的错误。任何想法来修复这个错误?谁能指出我为什么会发生此错误?有什么想法吗?

我的代码:

from sklearn.model_selection import KFold
from sklearn.metrics import mean_squared_error
from sklearn.linear_model import LinearRegression, Ridge, Lasso
from xgboost.sklearn import XGBRegressor

from sklearn.datasets import make_regression

models = [SVR(), RandomForestRegressor(), LinearRegression(), Ridge(), Lasso(), XGBRegressor()]
params = [{'C': [0.01, 1]}, {'n_estimators': [10, 20]}]

X, y = make_regression(n_samples=10000, n_features=20)

with warnings.catch_warnings():
    warnings.filterwarnings("ignore")
    cv = [[] for _ in range(len(models))]
    fold = KFold(5,shuffle=False)
    for tr, ts in fold.split(X):
        for i, (model, param) in enumerate(zip(models, params)):
            best_m = GridSearchCV(model, param)
            best_m.fit(X[tr], y[tr])
            s = mean_squared_error(y[ts], best_m.predict(X[ts]))
            cv[i].append(s)
    print(np.mean(cv, 1))

期望输出:

如果有办法解决上述错误,我希望选择带参数的最佳拟合模型,然后将其用于估计。任何改进上述尝试的想法?谢谢

当你定义

cv = [[] for _ in range(len(models))]

每个模型都有一个空列表。 但是,在循环中,您遍历了 enumerate(zip(models, params)),它只有 两个 元素,因为您的 params 列表有两个元素(因为 list(zip(x,y)) has length 等于 min(len(x),len(y))。

因此,您得到一个 IndexError,因为当您使用 np.mean.[=22 计算平均值时,cv 中的一些列表是空的(除前两个之外的所有列表) =]

解法: 如果您不需要在其余模型上使用 GridSearchCV,您可以使用空字典扩展 params 列表:

params = [{'C': [0.01, 1]}, {'n_estimators': [10, 20]}, {}, {}, {}, {}]

问题的根本原因是,当您要求评估 GridSearchCV 中的 6 个模型时,您只提供了前两个模型的参数:

models = [SVR(), RandomForestRegressor(), LinearRegression(), Ridge(), Lasso(), XGBRegressor()]
params = [{'C': [0.01, 1]}, {'n_estimators': [10, 20]}]

此设置下enumerate(zip(models, params))的结果,即:

for i, (model, param) in enumerate(zip(models, params)):
    print((model, param))

(SVR(C=1.0, cache_size=200, coef0=0.0, degree=3, epsilon=0.1, gamma='auto',
  kernel='rbf', max_iter=-1, shrinking=True, tol=0.001, verbose=False), {'C': [0.01, 1]})
(RandomForestRegressor(bootstrap=True, criterion='mse', max_depth=None,
           max_features='auto', max_leaf_nodes=None,
           min_impurity_decrease=0.0, min_impurity_split=None,
           min_samples_leaf=1, min_samples_split=2,
           min_weight_fraction_leaf=0.0, n_estimators=10, n_jobs=1,
           oob_score=False, random_state=None, verbose=0, warm_start=False), {'n_estimators': [10, 20]})

即最后 4 个模型被简单地忽略,所以你在 cv:

中得到它们的空条目
print(cv)
# result:
[[5950.6018771284835, 5987.293514740653, 6055.368320208183, 6099.316091619069, 6146.478702335218], [3625.3243553665975, 3301.3552182952058, 3404.3321983193728, 3521.5160621260898, 3561.254684271113], [], [], [], []]

试图获取 np.mean(cv, 1).

时导致下游错误

正如 Psi 在他们的回答中已经正确指出的那样,解决方案是在您实际上 执行任何 CV 搜索的模型中寻找空字典;省略了 XGBRegressor(还没有安装),结果如下:

models = [SVR(), RandomForestRegressor(), LinearRegression(), Ridge(), Lasso()]
params2 = [{'C': [0.01, 1]}, {'n_estimators': [10, 20]}, {}, {}, {}]

cv = [[] for _ in range(len(models))]
fold = KFold(5,shuffle=False)
for tr, ts in fold.split(X):
    for i, (model, param) in enumerate(zip(models, params2)):
        best_m = GridSearchCV(model, param)
        best_m.fit(X[tr], y[tr])
        s = mean_squared_error(y[ts], best_m.predict(X[ts]))
        cv[i].append(s)

其中 print(cv) 给出:

[[4048.660483326826, 3973.984055352062, 3847.7215568088545, 3907.0566348092684, 3820.0517432992765], [1037.9378737329769, 1025.237441119364, 1016.549294695313, 993.7083268195154, 963.8115632611381], [2.2948917095935095e-26, 1.971022007799432e-26, 4.1583774042712844e-26, 2.0229469068846665e-25, 1.9295075684919642e-26], [0.0003350178681602639, 0.0003297411022124562, 0.00030834076832371557, 0.0003355298330301431, 0.00032049282437794516], [10.372789356303688, 10.137748082073076, 10.136028304131141, 10.499159069700834, 9.80779910439471]]

print(np.mean(cv, 1)) 工作正常,给出:

[3.91949489e+03 1.00744890e+03 6.11665355e-26 3.25824479e-04
 1.01907048e+01]

因此,在您的情况下,您确实应该将 params 更改为:

params = [{'C': [0.01, 1]}, {'n_estimators': [10, 20]}, {}, {}, {}, {}]

正如 Psi 已经建议的那样。